Thank you for your answer, Remi.

But my question was not about momentum and was not about the classical limit. I want to know if within a description that is inherently quantum mechanical the velocity operator (NOT the momentum) i.e. 1/ih[H,x] may ever commute with the position operator. I already know that this is impossible for the standard Hamiltonian quadratic in the momentum. I wish to know if there are physically possible (massive particle) Hamiltonians for which this can happen.

Dear Aniello,

the problem is that you cannot separate velocity and momentum in non-relativistic QM (which is what you are looking at when you do not want to use the Dirac or Klein-Gordon equaltions). Let's also leave photons aside because they are, of course, never non-relativistic.

Then, as soon as you describe this particle by the Schrödinger equation, which is what you need to do in QM, you will have a quantised wave function, as imposed by the boundary conditions of the problem. And any such wave function is subject to Heisenberg's uncertainty relation.

Put more mathematically, you cannot write down any Hamiltonian without momentum, i.e., without a derivative with respect to the spatial coordinates. Thus, there will always be components which do not commute (of course some of them do, e.g., spatial x-direction and momentum in y-direction).

Given that momentum and velocity are inseparable for non-relativistic systems, that's the end of the story. One cannot go around this equivalence, even if one uses an alternative definition of the velocity operator.

In fact, you can show generally that:

i m [\hat H, \hat x] = \hbar \hat p,

where "\hat" indicates that whatever is written afterwards is an operator. You can even find this on Wikipedia, https://en.wikipedia.org/wiki/Ehrenfest_theorem, or in any QM textbook.

Best regards,

Alexander

It seems that higher order theories allow position and velocity to commute

http://www.isr.umd.edu/Labs/CSSL/simonlab/pubs/SimonPhysRevD1990.pdf

Dear Aniello,

I don't quite understand why you think they would commute in that paper. Three lines after Eq. (29), they state that {x,p} = xp + px =1, so they clearly do not commute.

Best regards,

Alexander

Dear Alexander,

Have a look at eq. (11), for the unconstrained Hamiltonian. For Lagrangians depending on, say, the acceleration, the corresponding Hamiltonian depends on (x,x',px,px'), where x' is the velocity. Since the latter is considered as an additional canonical variable, it commutes with x. 

Dear Aniello,

sorry, but I don't think you are correct. Eq. (11) tells us that position and momentum commute *IF THEY POINT IN ORTHOGONAL DIRECTIONS*. This is a trivial statement, which I already mentioned in my first reply to your qestion (quote: " Thus, there will always be components which do not commute (of course some of them do, e.g., spatial x-direction and momentum in y-direction)."). Nothing more than that is reflected by Eq. (11): once m=n, than position and momentum do not commute, as to be expected.

Once again: momentum is a derivative operator with respect to position, so they truly *CANNOT* commute. No derivative operator commutes with a non-constant function as long as it acts on the variables in that function. It's really as simple as that.

Best regards,

Alexander

Dear Alexander,

You are confusing momentum and velocity. It's trivial that momentum and position cannot commute, and indeed my question was about velocity, not momentum. 

Anyway, the q(m) are not orthogonal to each other. The motion is 1-dimensional. m is just the degree of the time derivative.

Dear Aniello,

after all we are back to square one... as I explained at length in my very first post, velocity and momentum are equivalent in non-relativistic QM.

If you don't accept that, then please go back to your textbooks and consult them. I have explained you my view and I am pretty sure that, no matter which active researcher you ask, you will get similar answers.

Best regards,

Alexander